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```ON THE COLOURS  OF  THICK   PLATES.                      185
25.    For the achromatic line jR = 0, and therefore s : c :: u : k. Hence s is to u in the constant ratio of c to k, and therefore, by a well-known geometrical theorem, the achromatic curve is a circle, having its centre in the line LQEQ produced, and cutting this line in the two points in which it is divided internally and externally in the given ratio.    The latter of these points may be formed by producing LE to meet L0EQ produced, and the former by producing LLQ till the produced part is equal to the line itself, and then joining the extremity of the produced part with E.    Hence the construction given in Art. 11 for determining the bright band of the order zero continues to hold good whatever be the angle of incidence.
26.    In the neighbourhood of the image the bands arc sensibly straight, being arcs of circles of very large radius.    To find the mean breadth of a band, it will be sufficient to suppose the point P to lie in the line L()E{}, to differentiate equation (33) making
R, \$, and u vary together, while ,v + u remains constant, replace /7 7? -\
7 by -=, and after differentiation take u and s to refer to the du J P'
small pencil, regarded as a ray, by which the image is seen. If i be the angle of incidence, we may put after differentiation s = c tan i, u = h tan i. We thus find
On account of the largeness of the angle of incidence, the breadth of the bands is sensibly uniform, and therefore ft may be regarded as the breadth of any one baud. It is to be remembered that ft denotes the linear breadth of a band as seen projected on the mirror. If we denote the angular breadth by -zzr, we have on account of the smallnoss of TO-.
__ f\$ cos i __     \c      V//,2 — sina i               ,^.
Ascci ~~t(c + A)     sin2if```